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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 1

In Cartesian components,
$$
\frac{\partial L}{\partial\dot r_i}=m\dot r_i+qA_i
$$
and
$$
\frac d{dt}\frac{\partial L}{\partial\dot r_i}
=m\ddot r_i+q\partial_tA_i+q\dot r_j\partial_jA_i,
\qquad
\frac{\partial L}{\partial r_i}
=-q\partial_i\phi+q\dot r_j\partial_iA_j.
$$
The <Euler-Lagrange equation> therefore gives
$$
m\ddot r_i
=q\left[-\partial_i\phi-\partial_tA_i
 +\dot r_j(\partial_iA_j-\partial_jA_i)\right].
$$
Using
$$
E=-\nabla\phi-\partial_tA,
\qquad B=\nabla\times A,
$$
and $[\dot r\times B]_i=\dot r_j(\partial_iA_j-\partial_jA_i)$, this becomes the <Lorentz force>
$$
\boxed{m\ddot r=q(E+\dot r\times B)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 8D / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 8D / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 8D / b / Solution

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